$\log_{9}81 = {?}$
If $\log_{b}x=y$ , then $b^y=x$ First, try to write $81$ , the number we are taking the logarithm of, as a power of $9$ , the base of the logarithm. $81$ can be expressed as $9\times9$ $81$ can be expressed as $9^2$ $9^2=81$, so $\log_{9}81=2$.